Better way for concatenating two sorted list of integers

Question

Lets assume I have one list and another tuple both of them are already sorted:

A = [10, 20, 30, 40]
B = (20, 60, 81, 90)

What I would need is to add all the elements from B into A in such a way that A remains sorted.

Solution I could come with was:

for item in B:
    for i in range(0, len(A)):
        if item > A[i]:
            i += 1
        else: 
            A.insert(i, item)

assuming A of size m, and B of size n; this solution would take O(mxn) in worst case, how can I make it perform better ?

Answer 1

A simple way would be heapq.merge:

A = [10, 20, 30, 40]

B = (20, 60, 81, 90)

from heapq import merge

for ele in merge(A,B):
    print(ele)

Output:

10
20
20
30
40
60
81
90

Some timings using the other O(n) solution:

In [53]: A = list(range(10000))

In [54]: B = list(range(1,20000,10))

In [55]: timeit list(merge(A,B))
100 loops, best of 3: 2.52 ms per loop

In [56]: %%timeit
C = []
i = j = 0
while i < len(A) and j < len(B):
    if A[i] < B[j]:
        C.append(A[i])
        i += 1
    else:
        C.append(B[j])
        j += 1
C += A[i:] + B[j:]
   ....: 
100 loops, best of 3: 4.29 ms per loop
In [58]: m =list(merge(A,B))
In [59]: m == C
Out[59]: True

If you wanted to roll your own this is a bit faster than merge:

def merger_try(a, b):
    if not a or not b:
        yield chain(a, b)
    iter_a, iter_b = iter(a), iter(b)
    prev_a, prev_b = next(iter_a), next(iter_b)
    while True:
        if prev_a >= prev_b:
            yield prev_b
            try:
                prev_b = next(iter_b)
            except StopIteration:
                yield prev_a
                break
        else:
            yield prev_a
            try:
                prev_a = next(iter_a)
            except StopIteration:
                yield prev_b
                break
    for ele in chain(iter_b, iter_a):
        yield ele

Some timings:

In [128]: timeit list(merge(A,B))
1 loops, best of 3: 771 ms per loop

In [129]: timeit list(merger_try(A,B))
1 loops, best of 3: 581 ms per loop

In [130]: list(merger_try(A,B))  == list(merge(A,B))
Out[130]: True

In [131]: %%timeit                                 
C = []
i = j = 0
while i < len(A) and j < len(B):
    if A[i] < B[j]:
        C.append(A[i])
        i += 1
    else:
        C.append(B[j])
        j += 1
C += A[i:] + B[j:]
   .....: 
1 loops, best of 3: 919 ms per loop

Answer 2

bisect module "provides support for maintaining a list in sorted order without having to sort the list after each insertion":

import bisect
for b in B:
    bisect.insort(A, b)

This solution does not create a new list.

---

Please note that bisect.insort(A, b) is equivalent to

A.insert(bisect.bisect_right(A, b), b)

Even though the search is fast (O(log n)), the insertion is slow (O(n)).

Answer 3

Lots of good discussion in this post! Arguing about timing is hard, so I wrote some timing script. It's quite rudimentary but I think it will do for now. I've attached the results too.

import timeit
import math
import matplotlib.pyplot as plt
from collections import defaultdict


setup = """
import bisect
import heapq
from random import randint


A = sorted((randint(1, 10000) for _ in range({})))
B = sorted((randint(1, 10000) for _ in range({})))


def bisect_sol(A, B):
    for b in B:
        bisect.insort(A, b)


def merge_sol(A, B):
    ia = ib = 0
    while ib < len(B):
        if ia < len(A) and A[ia] < B[ib]:
            if ia < len(A):
                ia += 1
        else:
            A.insert(ia + 1, B[ib])
            ib += 1


def heap_sol(A, B):
    return heapq.merge(A, B)


def sorted_sol(A, B):
    return sorted(A + B)
"""


sols = ["bisect", "merge", "heap", "sorted"]
times = defaultdict(list)
iters = [100, 1000, 2000, 5000, 10000, 20000, 50000, 100000]

for n in iters:
    for sol in sols:
        t = min(timeit.repeat(stmt="{}_sol(A, B)".format(sol), setup=setup.format(n, n), number=1, repeat=5))
        print("({}, {}) done".format(n, sol))
        times[sol].append(math.log(t))

for sol in sols:
    plt.plot(iters, times[sol])
plt.xlabel("iterations")
plt.ylabel("log time")
plt.legend(sols)
plt.show()

This is the result:

It's clear that modifying the list is the major bottleneck, so creating a new list is the way to go.

Answer 4

Here is a solution in O(n):

A = [10, 20, 30, 40]
B = [20, 60, 81, 90]
C = []
i = j = 0
while i < len(A) and j < len(B):
    if A[i] < B[j]:
        C.append(A[i])
        i += 1
    else:
        C.append(B[j])
        j += 1
C += A[i:] + B[j:]